Respuesta :
Answer:
Option E is correct, 79 μC
Explanation:
Given that,
A capacitor of capacitance C1
C1 = 3 μF
The capacitor is charge to 40V
V1=40V
Another capacitor of capacitance
C2 = 5 μF
Charge to 18V,
V2 =18V
When the positive terminal of one is connected to the negative terminal of the other, then, this is a series connection.
The capacitor are connected in series.
Series connection have the same charge.
Equivalent capacitance for series connection is give as
Ceq = C1 + C2
Ceq = 3μF + 5μF
Ceq = 8μF
Charge in a capacitor is given as
q = CV
Then, charge on Capacitor 1.
q1 = C1V1
q1 = 3μF × 40V
q1 = 120μC
Charge on capacitor 2
q2 = C2V2
q2 = 5μF × 18V
q2 = 90μC
The total charge in the circuit is
Qeq =q1 + q2
Qeq =90 + 120
Qeq = 210 μC
Then, the combine voltage in the circuit is
Qeq = CeqV
Then, V= Qeq/Ceq
V= 210 μC / 8 μC
V=26.25Volts
Then the charge on the 3μF is
q =CV
q =3 μC × 26.25 V
q = 78.75 μC
q ≈ 79 μC
Final answer is E.
The question is not typed properly! Complete question along with answer and step by step explanation is provided below.
Question:
A 3.0 μF capacitor is charged to 40 V and a 5.0 μF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other. What is the final charge on the 3.0 μF capacitor?
A) 11 μC
B) 15 μC
C) 19 μC
D) 26 μC
E) 79 μC
Given Information:
Capacitor 1 = C₁ = 3.0 μF
Capacitor 2 = C₂ = 5.0 μF
Initial voltage across capacitor 1 = V₁ = 40 V
Initial voltage across capacitor 2 = V₂ = 18 V
Required Information:
Charge on Capacitor 1 = q₁ = ?
Answer:
The correct option is A
Charge on Capacitor 1 ≈ 11 μC
Explanation:
Since the positive plate of one capacitor is connected to the negative plate of other capacitor then it means they are connected in series.
The change in charge is given by
Q = q₁ - q₂
Where q = CV
Q = C₁V₁ - C₂V₂
Q = 3*40 - 5*18
Q = 120 - 90
Q = 30 μC
Since capacitors are connected in series
ΔV₁ = ΔV₂
Since ΔV = q/C
q₁/C₁ = q₂/C₂
q₁/3 = q₂/5
q₁ = 3/5q₂
The total final charge is
Q = q₁' + q₂'
30 = 3/5q₂' + q₂ '
30 = 8/5q₂ '
q₂' = 30*5/8
q₂' = 18.75 μC
Therefore, the final charge on capacitor 1 is
q₁' = 3/5q₂'
q₁' = 3/5(18.75)
q₁' = 11.25 μC
That is close to option (A) 11 μC
q₁' ≈ 11 μC
